Problem: If y=x2+px+qy=x^{2}+p x+qy=x2+px+q, then if the least possible value of yyy is zero qqq is equal to.
Answer Choices:
A. 000
B. p24\dfrac{p^{2}}{4}4p2
C. p2\dfrac{p}{2}2p
D. −p2-\dfrac{p}{2}−2p
E. p24−q\dfrac{p^{2}}{4}-q4p2−q Solution:
The least value occurs when x=−p/2x=-p / 2x=−p/2. For x=−p/2x=-p / 2x=−p/2, y=(−p2/4)+q=0;∴q=p2/4y=\left(-p^{2} / 4\right)+q=0 ; \quad \therefore q=p^{2} / 4y=(−p2/4)+q=0;∴q=p2/4